Proposition 19

Let A be a stellar algebra and let I be a closed two-sided ideal of A. Then the quotient involutive Banach algebra $A/I$ is a stellar algebra.

The ideal I is self-adjoint (Lemma 15). Consider the stellar algebra $\widetilde{A}$ deduced from A by adjunction of a unit element (Def. 4 of I, p. 106). In this algebra, the set I is a closed self-adjoint two-sided ideal, and $A/I$ identifies with a closed involutive subalgebra of $\widetilde{A}/I$. One may therefore suppose that A is unital.

The Banach algebra $A/I$ is involutive. Let $\pi : A\rightarrow A/I$ be the canonical projection. The self-adjoint two-sided ideal I is a stellar subalgebra of A. Let $\mathfrak{F}$ be an increasing approximate unit of I (Prop. 18 of I, p. 121). Let us first show that for every $x\in A$, one has

(6) $\|\pi (x)\|_{A/I}=$ lim$_{f,\mathfrak{F}}\|x-xf\|$.

On the one hand, since $xf\in I$ for every $f\in I^{\leqslant 1}_+$, one has

$\|\pi (x)\|_{A/I}=$ inf$_{a\in I}\|x-a\|\leqslant$ lim inf$_{f,\mathfrak{F}}\|x-xf\|$.

On the other hand, for every $a\in I$, one has

and therefore, since $\|1-f\|\leqslant 1$ (Lemma 12 of I, p. 116) and $a\in I$, one deduces that

lim sup$_{f,\mathfrak{F}}\|x-xf\|\leqslant \|x-a\|$.

Thus it follows that lim sup$_{f,\mathfrak{F}}\|x-xf\|\leqslant \|\pi (x)\|_{A/I}$ since $a$ is arbitrary in I. Formula (6) is therefore proved.

Now let $x$ be an element of A. By formula (6), one has

$\|\pi (x)\|^2_{A/I}=$ lim$_{f,\mathfrak{F}}\|x-xf\|^2=$ lim$_{f,\mathfrak{F}}\|x(1-f)\|^2$

= lim$_{f,\mathfrak{F}}\|(1-f)x^*x(1-f)\|\leqslant$ lim$_{f,\mathfrak{F}}\|x^*x(1-f)\|=\|\pi (x^*x)\|_{A/I}$.

Lemma 6 of I, p. 103 then implies that the quotient involutive Banach algebra $A/I$ is a stellar algebra.